Find coordinates of a point Q on the graph $\sin (x) + \cos (y) = 0.5$ given that the gradient of its tangent is perpendicular to point P. Note:
Point $P$ is on the $y$-axis and above the $x$-axis
$\frac{-\pi}{6}\le x \le\frac{7\pi}{6}$
$\frac{-2\pi}{3}\le y\le\frac{2\pi}{3}$
What I have done so far:
Solving for $P$: 
$$x = 0
\\ \sin (0) + \cos (y) = 0.5
\\ 0 + \cos (y) = 0.5
\\ y= \pm\frac{\pi}{3} $$
For $P$, $y \gt 0$
$\therefore y = \frac{\pi}{3}$
Solving for $\frac{dy}{dx}$: 
$$\sin(x) + \cos(y) = 0.5
\\ \cos(x) - \sin(y)\frac{dy}{dx} = 0$$
$\therefore \frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}$
Derivative at $P$:
$$\frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}
= \frac{\cos(0)}{\sin(\frac{\pi}{3})}
= \frac{2}{\sqrt3}$$
As for the gradient of the tangent line at $Q$ is perpendicular to that at $P$:
$\frac{dy}{dx} = \frac{-\sqrt3}{2}$
How do I solve for the coordinates of $Q$ after this?
 A: Good work. Note the point $Q$ lies on the curve as well as the tangent line. So:
$$\begin{cases}\frac{\cos x}{\sin y}=-\frac{\sqrt{3}}{2}\\ \sin x+\cos y=0.5 \end{cases} \Rightarrow$$
From the first:
$$\cos^2x=\frac34(1-\cos^2y)\Rightarrow \cos y=\pm\sqrt{1-\frac43\cos^2x}$$
Now sub it to the second:
$$1-\frac43\cos^2x=\frac14-\sin x+\sin^2x\Rightarrow \\
4\sin^2x+12\sin x-7=0 \Rightarrow \sin x=\frac12.$$
Referring to the given constraints, the final answer is:
$$x=\pi-\frac{\pi}{6},y=\frac{\pi}{2}$$
A: Note that the curve (one of an infinite number of such "loops" satisfying the given equation) is symmetrical about the point $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ ; $  this is a consequence of the trigonometric function properties $ \ \cos (-y) \ = \ \cos y \ $ and $ \ \sin \left(\frac{\pi}{2} + \xi \right) \ = \ \sin \left(\frac{\pi}{2} - \xi \right) \ \ . $  If we use this latter symmetry to define a coordinate $ \ x \ = \ \frac{\pi}{2} + \xi \ \Rightarrow \ \xi \ = \ x  -  \frac{\pi}{2} \ \ , $ then the curve equation becomes $ \ \sin \left(\frac{\pi}{2} + \xi \right) + \cos y \ = \ 0.5 \ $ over the intervals $ \ \frac{-2\pi}{3} \ \le \ \xi \ , \ y \ \le \ \frac{2\pi}{3} \ \ . $  The slope of the tangent line at $ \ P(x,y) \ $ is then
$$ \frac{dy}{dx}|_P \ \  = \ \ \frac{dy}{d \xi}|_P \ \  = \ \  \frac{\cos\left(\frac{\pi}{2} + \xi_P \right)}{\sin (y_P)}  $$
and the slope of the normal line (which is equal to the slope of the tangent line at $ \ Q \ $ ) is
$$ \frac{dy}{dx}|_Q \ \  = \ \   -\frac{\sin (y_P)}{\cos\left(\frac{\pi}{2} + \xi_P \right)} \ \ =  \ \ \frac{\cos\left(\frac{\pi}{2} + \xi_Q \right)}{\sin (y_Q)} \ \ , $$
The "angle-addition" formula for cosine yields $ \ \cos\left(\frac{\pi}{2} + \xi_Q \right) \ = \ \cos\left(\frac{\pi}{2}  \right)·\cos \xi_Q \ - \ \sin\left(\frac{\pi}{2}\right) ·\sin \xi_Q $ $ = \ -\sin \xi_Q \ \ , $ and similarly, $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ -\sin y_Q \ \ . $  Thus, we may write
$$ \frac{dy}{dx}|_Q \ \  = \ \  \frac{-\sin \xi_Q}{-\cos\left(\frac{\pi}{2} + y_Q \right)} \ \ = \ \ \frac{\sin \xi_Q}{\cos\left(\frac{\pi}{2} + y_Q \right)} \ \ = \ \ -\frac{\sin (y_P)}{\cos\left(\frac{\pi}{2} + \xi_P \right)} \ \ , $$
which now suggests how to make correspondences between coordinates.
You identified $ \ P \ $ as one of the $ \ y-$intercepts $ \ (x_P \ = \ 0 \Rightarrow \ \xi_P \ = \ -\frac{\pi}{2} ) \ $ of the curve, which has  $ \ \cos(y_P) \ = \ 0.5 \ \Rightarrow \ y_P \ = \ \frac{\pi}{3} \ \ . $  We may then take
$$ \sin(y_P) \ \ = \ \ \frac{\sqrt3}{2} \ \ = \  \ \sin(\xi_Q) \ \ \ , \ \ \ \cos\left(\frac{\pi}{2} + \xi_P \right) \ \ = \ \ 1 \ \ = \ \ -\cos\left(\frac{\pi}{2} +  y_Q \right) \ \ , $$
or
$$ \sin(y_P) \ \ = \ \ \frac{\sqrt3}{2} \ \ = \  \ -\sin(\xi_Q) \ \ \ , \ \ \ \cos\left(\frac{\pi}{2} + \xi_P \right) \ \ = \ \ 1 \ \ = \ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ \ . $$
The first pair of equations gives us $ \ \sin(\xi_Q) \ = \ \frac{\sqrt3}{2} \ \Rightarrow \ \xi_Q \ = \ \frac{\pi}{3} \ , \ \frac{2\pi}{3} \ \Rightarrow \ x_Q \ = \ \frac{5\pi}{6} \ , \ \frac{7\pi}{6} \ $ and $ \ \cos\left(\frac{\pi}{2} +  y_Q \right) \ = \ -1 \ \Rightarrow \ \frac{\pi}{2} +  y_Q \ = \  \pi \ \Rightarrow \ y_Q \ = \ \frac{\pi}{2} \ \ ; $ of these two points, $ \ \sin \left( \frac{5\pi}{6} \right) \ + \ \cos \left( \frac{\pi}{2} \right) \ = \ 0.5 \   $ satisfies the curve equation.  On the other hand, the second pair produces $ \ \sin(\xi_Q) \ = \ -\frac{\sqrt3}{2} $ $ \Rightarrow \ \xi_Q \ = \ -\frac{\pi}{3} \ , \ -\frac{2\pi}{3} \ \Rightarrow \ x_Q \ = \ \frac{\pi}{6} \ , \ -\frac{\pi}{6} \ $ and $ \ \cos\left(\frac{\pi}{2} +  y_Q \right) \ = \ 1 \ \Rightarrow \ \frac{\pi}{2} +  y_Q \ = \  0 \ \Rightarrow \ y_Q \ = \ -\frac{\pi}{2} \ \ , $ for which only $ \ \sin \left( \frac{\pi}{6} \right) \ + \ \cos \left( -\frac{\pi}{2} \right) \ = \ 0.5 \   $ works in the equation.
This tells us that the location of point $ \ Q \ $ is $ \ \left(\frac{5\pi}{6} \ , \ \frac{\pi}{2} \right) \ \ . $  We have also found, as we would expect from the symmetry of this loop about $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ , $ that there is a second point $ \ Q' \ \left(\frac{\pi}{6} \ , \ \frac{-\pi}{2} \right) \   $ at which the tangent line is perpendicular to the tangent line at $ \ P \ \ . $
By a related argument, we can also show that there is a point $ \ P' \ \left( \pi  \ , \ \frac{-\pi}{3} \right) \   $ with a tangent line parallel to the one at $ \ P \ \ .  $  Naturally, because the full "curve" described by the equation covers the plane with "loops" with periodicity $ \ 2 \pi \ $ in the $ \ x-$ and $ \ y-$directions, there are "families" of points $ \ \mathcal{P}   \ \left(2 m \pi \ , \ \frac{\pi}{3} + 2 n \pi \right) \ \   $ and  $ \ \mathcal{Q}   \ \left(\frac{5\pi}{6} + 2 m \pi \ , \ \frac{\pi}{2} + 2 n \pi \right) \ \ , \ m \ \ \text{and} \ \ n \  $ being integers, which have the prescribed relationship.

